# Optimal idle or frame sync symbols for OQPSK

Okay, everyone here knows that my thing was audio and that I have never gotten paid to do anything regarding communications systems. So, while I know a thing or two about DSP and even a little about discrete mathematics (from which we get GF2 or Maximum Length Sequences) I am very humble about my "expertise" in this area.

But OQPSK really intrigues me because it appears to me to be very elegant. (And people here know that I am in love with elegance in signal processing.) I have posted here a couple times. And once previously with a question and once over at the EE stack exchange.

Now, for the purpose of illustration, I am assuming we have a single bit stream, $$a[n] \in \{$$0,1$$\}$$, that becomes a bipolar bit stream

\begin{align} x[n] &\triangleq -1 + 2 a[n] \quad \in \{-1, 1 \} \\ &= -(-1)^{a[n]} \\ \end{align}

Now, part of what makes OQPSK so elegant is that the bits can be encoded onto bipolar quadrature signals $$i[n]$$ and $$q[n]$$ quite naturally where the even bits go into $$i[n]$$ and the odd bits go into $$q[n]$$. No delay is necessary, and no "dibit" symbols need to be formed as you would with non-offset QPSK.

\begin{align} i[n] \ &= \ g[n] x[n] \ + \ (1-g[n]) x[n-1] \\ q[n] \ &= \ (1-g[n]) x[n] \ + \ g[n] x[n-1] \\ \end{align}

where $$g[n]$$ is an even/odd gating signal defined as

$$g[n] \triangleq \tfrac{1}{2}\left( 1 + (-1)^n \right)$$

and

$$1-g[n] = \tfrac{1}{2}\left( 1 - (-1)^n \right)$$

Note that for $$n$$ even, $$g[n]=1$$ and only $$i[n]$$ can change, while for $$n$$ odd, $$1-g[n]=1$$ and only $$q[n]$$ can change.

Now this is equivalent to having a rectangular pulse as the continuous-time symbol transmitted with each bit (assume the sampling period is the same as the bit period):

\begin{align} i(t) &= \sum\limits_{n=-\infty}^{\infty} x[2n] \, p(t-2n) \\ \\ q(t) &= \sum\limits_{n=-\infty}^{\infty} x[2n+1] \, p(t-(2n+1)) \\ \end{align}

where

$$p(t) \triangleq \begin{cases} 0 \qquad & t < 0 \\ 1 \qquad & \qquad 0 \le t < 2 \\ 0 \qquad & \qquad \qquad \qquad 2 \le t \\ \end{cases}$$

But, if we're gonna contain the channel bandwidth, then a better symbol pulse is:

$$p(t) = \operatorname{sinc}\left( \tfrac{t}{2} \right) w(t)$$

where

$$\operatorname{sinc}(u) \triangleq \begin{cases} \frac{\sin(\pi u)}{\pi u} \qquad & u \ne 0 \\ 1 & u = 0 \\ \end{cases}$$

and $$w(t)$$ is a good window. It could be a Hamming window

$$w(t) \triangleq \begin{cases} \tfrac{27}{50} + \tfrac{23}{50} \cos\left(2\pi \tfrac{t}{M}\right) \quad \quad & |t| \le \tfrac{M}{2} \\ 0 & |t| > \tfrac{M}{2} \\ \end{cases}$$

but, I still am a partisan for the Kaiser window in this application:

$$w(t) \triangleq \begin{cases} \frac{1}{J_0(\beta)} J_0\left(\beta \sqrt{1 - \left(\frac{t}{1+M/2}\right)^2} \right) \quad \quad & |t| \le \tfrac{M}{2} \\ 0 & |t| > \tfrac{M}{2} \\ \end{cases}$$

$$J_0(\cdot)$$ is the 0th-order modified Bessel function of the first kind.

$$J_0(u) = 1 \ + \ \sum\limits_{k=1}^{\infty} \frac{1}{(k!)^2} \left(-\frac{u^2}{4}\right)^{k}$$

$$M+1$$ is the number of non-zero samples or FIR taps. $$\beta$$ is a "shape parameter" and O&S recommend this heuristic:

$$\beta = \begin{cases} 0.1102 \cdot (A-8.7) & A>50 \\ 0.5842 \cdot (A-21)^{2/5} + 0.07886 \cdot (A-21) \quad & 21 < A \le 50 \\ 0.0 & A \le 21 \\ \end{cases}$$

$$M = 2 \left\lceil \frac{A-8}{4.57 \cdot \Delta\omega} \right\rceil$$

$$A$$ is the desired stopband attention in dB and $$\Delta\omega$$ is the desired width of the transition band in normalized angular frequency. $$\lceil \cdot \rceil$$ is the ceiling function (i.e. "round up").

Then this becomes:

\begin{align} i(t) &= \sum\limits_{n=\lfloor t/2-M/4 \rfloor}^{\lfloor t/2+M/4 \rfloor} x[2n] \, \operatorname{sinc}\left( \tfrac{t-2n}{2} \right) w(t-2n) \\ \\ q(t) &= \sum\limits_{n=\lfloor (t-1)/2-M/4 \rfloor}^{\lfloor (t-1)/2+M/4 \rfloor} x[2n+1] \, \operatorname{sinc}\left( \tfrac{t-(2n+1)}{2} \right) w(t-(2n+1)) \\ \end{align}

$$\lfloor \cdot \rfloor$$ is the floor() function (always round down).

This has no ISI for either $$i(t)$$ or $$q(t)$$ (assuming the receiver is properly time-aligned, which is why synchronization is salient) and it occupies the entire band but (if the window is good) very little energy outside of the band. I am conviced that this "Nyquist FDM" is the right way to do this.

Now we can tell that any of these four sequences will not move the quadrature $$i(t)\,+\,j\,q(t)$$ vector around:

00000000000000000000000000000000

11111111111111111111111111111111

01010101010101010101010101010101

10101010101010101010101010101010

and that this sequence will send our quadrature vector spinning counter-clockwise at the maximum rotational speed:

00110011001100110011001100110011

and this in the maximum clockwise rotational speed:

01100110011001100110011001100110

Now I used to be saying that these would make good frame-sync symbols (and maybe idle channel signals) because they end up being two pure frequencies that are at the maximum and minimum frequencies in the band. I no longer think that is optimal because if the bands are packed like the Nyquist FDM (in the middle), I don't want heavy carriers hanging out at the bandedge possibly disturbing neighboring bands.

If the channel is idling, I now think it should be in the middle of the band rather than at the edge.

So now my question is, for an idle channel, what should be the bit pattern of the idle channel? One of those first four streams above (so that it is a constant position of the vector and rock-solid on the center frequency)? Or some other bit pattern that has just as many clockwise rotations as counter-clockwise rotations (so that the mean phase change is zero and the center frequency is zero and constant)? Should an idle channel occupy the Nyquist FDM bandwidth (nearly) completely? Or should it be only the center frequency.

And then my next question will be about Barker codes and using that for the "start bit" or the frame sync word.

• So here is a comment about lack of elegance in your understanding of OQPSK systems (and digital communication systems too). The elegant mapping from $a[n]\in\{0,1\}$ to $x[n]\in\{+1,-1\}$ is $x[n]=(-1)^{a[n]}$ which sends $0\to 1, 1\to -1$. Big deal! you say? Not so. Since you have identified $\{0,1\}$ as the field GF$(2)$, the mapping I have suggested transfers the additive structure of GF$(2)$ to the multiplicative structure of the roots of unity. Dec 7, 2021 at 21:20
• @DilipSarwate , I totally agree. The reason I used the other convention is because of the diagram that i lifted from Wikipedia. If you look at the little MLS tutorial I did on comp.dsp that also lives at dspguru.com you'll see I use the better convention. Dec 8, 2021 at 2:14

Same thing with the barker code such as my favorite to explain correlation with sequences so I know it by heart: 10110111000. That will add (in GF2) to 11 only when they are aligned and -1 everywhere else when we send that sequence repeatably. So this offers an alignment with processing gain on the aligned result ($$10\log_{10}(11)$$ dB increase in SNR when each sample is uncorrelated/independent) within that degree of granularity over the individual bit patterns. You'll see sequences in headers that are combinations of all of these features, combining PRN sequences with 101010101 patterns.
• So Dan (it was nice to meet with you on Friday), let's say we do the GF2 (a.k.a. "MLS") thing for the idle channel. Does that make sense? Those Maximum Length Sequences have length of $2^N-1$, not $2^N$. Now, I am not committed to words in the serial stream having to be 32 or 64 bits, but since the $i[n]$ and $q[n]$ are sorta separate and independent, it seems to me that maybe the word length for both could be 31 bits, so that a length 31 MLS will have that strong spike every 62 bit periods. And the correlation I am thinking about would be with the sinc functions and the MLS. Dec 5, 2021 at 23:49
• I remember finding a more exhaustive list, but this list of primitive polynomials looks pretty good. There are only 3 for 31 bit sequences ($2^5 - 1$) but I think that the bit reversed polynomials are also primitive, just not unique in symmetry. My idea is that the idle channel should be a repeating sequence of 31 bits and that's just for $i[n]$ and then $q[n]$ can be a different 31-bit repeating sequence (possibly the reverse sequence) and the two will light up a matched filter good when the 62-bit word is aligned. Dec 6, 2021 at 0:45